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1989 AIME Problems/Problem 14

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Problem

Given a positive integer n, it can be shown that every complex number of the form r+si, where r and s are integers, can be uniquely expressed in the base -n+i using the integers 1,2,\ldots,n^2 as digits. That is, the equation

r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0

is true for a unique choice of non-negative integer m and digits a_0,a_1,\ldots,a_m chosen from the set \{0,1,2,\ldots,n^2\}, with a_m\ne 0. We write

r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}

to denote the base -n+i expansion of r+si. There are only finitely many integers k+0i that have four-digit expansions

k=(a_3a_2a_1a_0)_{-3+i}~~~~a_3\ne 0.

Find the sum of all such k.

Solution

First, we find the first three powers of -3+i:

(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i

So we need to solve the diophantine equation a_1-6a_2+26a_3=0 \Longrightarrow a_1-6a_2=-26a_3.

The minimum the left hand side can go is -54, so a_3\leq 2, so we try cases:

  • Case 1: a_3=2
The only solution to that is (a_1, a_2, a_3)=(2,9,2).
  • Case 2: a_3=1
The only solution to that is (a_1, a_2, a_3)=(4,5,1).
  • Case 3: a_3=0
a_3 cannot be 0, or else we do not have a four digit number.

So we have the four digit integers (292a_0)_{-3+i} and (154a_0)_{-3+i}, and we need to find the sum of all integers k that can be expressed by one of those.

(292a_0)_{-3+i}:

We plug the first three digits into base 10 to get 30+a_0. The sum of the integers k in that form is 345.

(154a_0)_{-3+i}:

We plug the first three digits into base 10 to get 10+a_0. The sum of the integers k in that form is 145. The answer is 345+145=\boxed{490}.

See also

1989 AIME (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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